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Part CCSIR NET June 2023p-squared-groups-abelian-not-cyclic

P squared groups abelian not cyclic

Let G be a group of order 2023. Which of the following statements are true?

A.G is an Abelian group.✓
B.G is a cyclic group.
C.G is a simple group.
D.G is not a simple group.✓

Solution

$2023 = 7 \cdot 1 7^{2} . n_{7} \equiv 1 (m o d 7)$ and $n_{7}$ | $289 \Rightarrow n_{7} = 1; n_{17}$ | 7 and $\equiv 1 (m o d 17) \Rightarrow n_{17} = 1$ . So $G ≅ Z_{7} \times P$ with P of order $1 7^{2}$ abelian: G is abelian and not simple. P may be $Z_{17} \times Z_{17}$ , so G need not be cyclic.

The trap it tests

Invariants don't determine

Two objects sharing an invariant were treated as the same object.

Drill statements like this

Related counterexample: A group of order pq (p < q) is always cyclic

From Groups › Sylow theorems and groups of small order

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